Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 20

Find the derivative of $\displaystyle f(x) = 1.5x^2 - x + 3.7$ using the definition and the domain of its derivative.

Using the definition of derivative


$
\begin{equation}
\begin{aligned}

\qquad f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
&&
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{1.5(x + h)^2 - (x + h) + 3.7 - (1.5x^2 - x + 3.7)}{h}
&& \text{Substitute $f(x + h)$ and $f(x)$}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\cancel{1.5x^2} + 3xh + 1.5h^2 - \cancel{x} - h + \cancel{3.7} - \cancel{1.5}x^2 + \cancel{x} - \cancel{3.7}}{h}
&& \text{Expand and combine like terms}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{3xh + 1.5h^2 - h}{h}
&& \text{Factor the numerator}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\cancel{h}(3x + 1.5h - 1)}{\cancel{h}}
&& \text{Cancel out like terms}
\\
\\
f'(t) &= \lim_{h \to 0} 3x + 1.5h - 1 = 3x + 1.5(0) - 1
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


$\qquad \fbox{$f'(x) = 3x - 1$}$

$f(x)$ is a polynomial function while $f'(x)$ is a linear function. Both of the functions are continuous in every number. Therefore, their domain is $(-\infty, \infty)$